Solve y''-4y'+4y=3e^2t

Question

Solve. y'' - 4y' + 4y = 3e^2t

Answer 1

First, solve the homogenous equation y''- 4y + 4y = 0. The corresponding characteristic equation is

r²-4r + 4 =0, or (r-2)²=0. Then the homogeneous solution is y_h= C₁ e^2t + C₂ t e^2t.

Now, let's use the method of variation of parameters to find a particular solution. That is, we seek a solution of the form y_p = u e^2t + v t e^2t, where u and v satisfy the equations

u =- ∫(t e^2t )(3 e^2t)/W and v = ∫(e^2t )(3 e^2t)/W.

Here W is the Wronskian of the homogeneous solutions which can be computed as the determinant of

e^2t te^2t

2e^2t e^2t + 2t e^2t

so W = e^4t + 2te^4t - 2te^4t = e^4t.

Then u =- ∫3t = -3t²/2 , and v = ∫3 = 3t.

So y_p = -3t²e^2t/2 + 3t²e^2t = 3t²e^2t/2.

Altogether, the general solution is y = C₁ e^2t + C₂ t e^2t + 3t²e^2t/2.